# Question #3acdc

##### 1 Answer

#### Answer:

Here's how you can do that.

#### Explanation:

The idea here is that you need to work with a sample of this solution and figure out how many moles of propanol it contains.

As you know, **molality** is defined as the number of moles of solute present **for every** **of solvent**. In your case, propanol is the solute and water is the solvent.

To make the calculations easier, let's pick a sample of solution that contains exactly

#1 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 10^3"g"#

of **water**. You know that water has a **molar mass** of

#10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"#

Now, the **mole fraction** of propanol in this solution is defined as the ratio between the number of moles of propanol, let's say **total number of moles** present in solution.

In your case, this mole fraction is equal to

#(n_p color(red)(cancel(color(black)("moles"))))/((n_p + 55.51)color(red)(cancel(color(black)("moles")))) = 0.1538#

This means that you have

#n_p = (n_p + 55.51) * 0.1538#

#(1 - 0.1538) * n_p = 55.51 * 0.1538#

#0.8462 * n_p = 8.5374 implies n_p = 8.5374/0.8462 = 10.09#

Therefore, you know that a solution of propanol that contains **moles** of propanol.

You can thus say that the molality of the solution is equal to

#color(darkgreen)(ul(color(black)("molality = 10.09 mol kg"^(-1))))#

The answer is rounded to four **sig figs**, the number of sig figs you have for the mole fraction of propanol.